A glass ball collides with a smooth horizontal surface (`xz` plane) with a velocity `V = ai- bj`. If the coefficient of restitution of collision be `e`, the velocity of the ball just after the collision will be

Options

A. `sqrt(e^(2)a^(2)+b^(2))` at angle `tan^(-1)(a/(eb))` to the vertical
B. `sqrt(a^(2)+e^(2)b^(2))` at angle `tan^(-1)(a/(eb))` to the vertical
C. `sqrt(a^2+b^2/e^2)` at angle `tan^-1((ea)/b)` to the vertical
D. `sqrt(a^(2)/e^(2)+b^(2))` at angle `tan^(-1)(a/(eb))` to the vertical

Explanation

To solve the problem of determining the velocity of a glass ball just after it collides with a smooth horizontal surface, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Velocity**:
The initial velocity of the ball is given as \( \mathbf{V} = a \hat{i} - b \hat{j} \), where \( a \) is the component in the x-direction and \( -b \) is the component in the y-direction.

2. **Understand the Collision**:
The collision occurs with a smooth horizontal surface (the xz-plane). The normal to this surface is in the y-direction. Therefore, the y-component of the velocity will be affected by the collision, while the x-component will remain unchanged.

3. **Apply the Coefficient of Restitution**:
The coefficient of restitution \( e \) is defined as the ratio of the velocity of separation to the velocity of approach. For a collision with a surface, the velocity of approach is the negative of the y-component of the incoming velocity, which is \( b \).

\[ e = \frac{\text{Velocity of separation}}{\text{Velocity of approach}} = \frac{v_{y, \text{after}}}{b} \]

4. **Determine the Velocity of Separation**:
From the equation above, we can express the velocity of separation in the y-direction after the collision:

\[ v_{y, \text{after}} = e \cdot b \]

Since the ball is moving upwards after the collision, the direction of this component will be positive.

5. **Combine the Components of Velocity After Collision**:
The final velocity of the ball after the collision can be expressed as:

\[ \mathbf{V}_{\text{final}} = a \hat{i} + e b \hat{j} \]

Here, \( a \hat{i} \) remains unchanged, and \( e b \hat{j} \) is the new y-component of the velocity.

6. **Conclusion**:
The velocity of the ball just after the collision is:

\[ \mathbf{V}_{\text{final}} = a \hat{i} + e b \hat{j} \]